Let $R$ be the region enclosed by the line $y=\dfrac x3$, the line $y=4-x$, and the $x$ -axis. $y$ $x$ ${y=4-x}$ ${y=\dfrac x3}$ $x=5}$ $ 0$ $(3,1)$ $(4,0)$ $ R$ A solid is generated by rotating $R$ about the line $x=5$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=4-x}$ ${y=\dfrac x3}$ $x=5}$ $ 0$ $(3,1)$ $(4,0)$ Let the thickness of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ This is called the washer method. What we now need is to figure out the expressions of $r_1(y)$ and $r_2(y)$, and the interval of integration. $r_1(y)$ is equal to the distance between the line $y=\dfrac x3$ and the line $x=5$. To find it, we need to solve the equation for $x$ : $x=3y$ So, ${r_1(y)=5-3y}$. $r_2(y)$ is equal to the distance between the line $y=4-x$ and the line $x=5$. To find it, we need to solve the equation for $x$ : $x=4-y$ So, ${r_2(y)=y+1}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [({5-3y})^2-({y+1})^2] \\\\ &=\pi\left[ 25-30y+9y^2-(y^2+2y+1) \right] \\\\ &=\pi(8y^2-32y+24) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=1$. So the interval of integration is $[0,1]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^1 \left[\pi(8y^2-32y+24) \right] dy \\\\ &=\pi\int_0^1 (8y^2-32y+24)dy \end{aligned}$ Let's evaluate the integral. $\pi\int_0^1 (8y^2-32y+24)dy=\dfrac{32\pi}{3}$ In conclusion, the volume of the solid is $\dfrac{32\pi}{3}$.